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too complicated to be written here. Click on the link to download a text file. |
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X(2), X(6), X(32), X(1501), X(9233), X(3504), X(18898), X(18899), X(41293), X(41294), X(41295), X(41309) |
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Geometric properties : |
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K1182 is the locus of pole of pK(Ω, P) passing through X(2), X(6), X(32). K1182 is a nodal isocubic with node X(32). Its root is the barycentric product X(110) x X(385), the intersection of the lines {6,22},{110,351}. • When Ω = X(32), the pivot P must lie on the line (L) passing through X(2), X(32) and many other centers. (L) is the trilinear polar of X(4577). These pK(X32, P) are in a same pencil which contains K177 = pK(X32, X2). See the related Table 34. • When Ω ≠ X(32), there is one and only one such pK with pole Ω and then its pivot P must lie on the circum-conic (C) with perspector X(5027). (C) passes through X(6), X(385), X(1691), X(3407) corresponding to the cubics K1016, K128, K789, K1013 respectively. K1182 passes through every center Ω(n) with first barycentric : a^6(a^2b^2 + a^2c^2 + b^2c^2 n) / (b^2 + c^2 + a^2 n) where n can be a real number or infinity and more generally any rational symmetric function f(a,b,c) with numerator and denominator of same degree. With Ω(n) ≠ X(32), the pivot on (C) of the pK has first barycentric : a^2 (a^4 - b^2 c^2) / (b^2 + c^2 + a^2 n). For example Ω(∞) = X(6), Ω(0) = X(9233), Ω(1) = X(1501). With n = – (a^2 + b^2) (a^2 + c^2) (b^2 + c^2) / (a^2 b^2 c^2), we find Ω(n) = X(2). Obviously, two special different values of n give the same point X(32), the node of the cubic. These are : 1 ± √(a^2 + b^2 + c^2) √(a^2 b^2 + a^2 c^2 + b^2 c^2) / (a b c). See K1183 for other related properties and more points on K1182. |